What is a reasonable measurement for the distance to Neptune?
2 answers:
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Answer:
13.309 m/s²
Explanation:
Length from shoulder to hand, l = 30 cm = 0.3 m
initial velocity, u = 1 m/s
final velocity, v = 2.5 m/s
time, t = 3 s
Let the tangential acceleration is a.
by using first equation of motion
v = u + at
2.5 = 1 + 3 a
a = 0.5 m/s²
Let the centripetal acceleration is a'.
a' = v'²/l
a' = 2 x 2 / 0.3
a' = 13.3 m/s²
The tangential acceleration and the centripetal acceleration are both perpendicular to each other. So, the net acceleration is given by
A = 13.309 m/s²